What is the reduction potential at 298 K of a half-cell consisting of a Pt electrode dipped in a solution that is 1.5 M in Fe2+ and 0.015 M in Fe3+? (E°Fe3+/Fe2+ = 0.770 V; 2.303RT/F = 0.059 V)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.652 V
The half-cell reaction is \( \ce{Fe^3+ + e^- -> Fe^2+} \), with n = 1. The Nernst equation for its reduction potential is:
\( \displaystyle E=E^{\circ}-\frac{2.303RT}{nF}\log\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} \)
\( \displaystyle E=0.770\ \mathrm{V}-\frac{0.059\ \mathrm{V}}{1}\log\frac{1.5}{0.015} \)
\( \displaystyle E=0.770\ \mathrm{V}-0.059\ \mathrm{V}\times\log 100=0.770\ \mathrm{V}-0.118\ \mathrm{V} \)
\( \displaystyle E=0.652\ \mathrm{V} \)
The reduced form \( \ce{Fe^2+} \) is 100 times more concentrated than \( \ce{Fe^3+} \), so the reduction potential is lower than E°.